 Solution

For independent bond percolation on ${\mathbb{Z}}^d$, $d\ge 2$, there is almost surely at most one infinite percolation cluster at every parameter. This is the Burton-Keane theorem. It does not require an assumption about the behavior at the critical probability. If the probability that the origin lies in an infinite percolation cluster is positive, translation ergodicity of Bernoulli percolation also gives almost-sure existence and hence exactly one.

The two structural ingredients are translation invariance and finite-energy property of Bernoulli percolation. For $0<p<1$, every prescribed configuration of finitely many bonds has positive conditional probability given all the other bonds. In particular, opening or closing a fixed finite set of bonds transforms events of positive probability into events of positive probability. The product measure is ergodic under translations. One way to see this is to approximate a translation-invariant event by a finite-cylinder event, translate the cylinder sufficiently far to make its coordinates disjoint, and use independence. Sending the approximation error to zero gives $\mathbb{P}(A)=\mathbb{P}(A{)}^2$.

Let $K$ be the number of infinite percolation clusters. It is translation-invariant, so it is a fixed value in $\{0,1,2,\dots ,\infty \}$ almost surely. First exclude a finite value $k\ge 2$. Some finite box meets at least two infinite percolation clusters with positive probability. Force every internal bond in that box to be open. This joins those clusters and can only merge other clusters; it cannot split an infinite percolation cluster. It also cannot create a new infinite percolation cluster out of finite ones, since only finitely many clusters touch the modified region. Thus the number of infinite percolation clusters decreases. finite-energy property of Bernoulli percolation makes the modified event have positive probability, contradicting its almost-sure value $k$.

Suppose next that $K=\infty$. Some finite box meets three different infinite percolation clusters with positive probability. Each such cluster has an infinite component outside the box: deletion of a finite set in a locally finite graph leaves only finitely many components adjacent to that set, so an infinite percolation cluster cannot have all of them finite. Choose one infinite exterior component from each of the three clusters, and one bond connecting it to the box.

Inside the box, connect the three selected exterior endpoints by a finite tree, preserving just their three selected entrance bonds. Take the minimal subtree connecting these endpoints; they are leaves, since their only retained connection into the box is their chosen entrance bond. This tree has a branch graph vertex of degree three. Close every other internal or boundary bond of the box. Its three exterior components remain distinct outside, since they belonged to different original clusters. The branch graph vertex now has exactly three open incident bonds, and deleting it separates its cluster into three infinite components. Such a graph vertex is a trifurcation vertex in percolation. There are finitely many choices for the endpoints and tree within the chosen region, so at least one fixed finite modification of Bernoulli percolation has positive probability. finite-energy property of Bernoulli percolation consequently gives positive probability of a trifurcation at some fixed graph vertex. Translation invariance makes this probability the same number $q>0$ at every graph vertex.

The decisive fact is that a finite box cannot contain a positive-volume density of these branching points. Here is the boundary count, including its combinatorial justification. For a finite graph vertex set $W$, let ${\partial }^{+}W$ be the exterior graph vertices adjacent to it. In each infinite percolation cluster meeting $W$, contract each of its components outside $W$ that is adjacent to $W$ to a terminal graph vertex. The quotient graph consisting of the cluster's graph vertices in $W$ and these terminals is finite and connected. A trifurcation graph vertex in $W$ separates its terminals into at least three nonempty groups, since each of its infinite branches must leave $W$.

Take a minimal subtree of the quotient graph connecting all the terminals. Every such trifurcation graph vertex must belong to that subtree and have degree at least three: otherwise the terminals in one of its separated groups could not connect to the other groups. All leaves of the minimal subtree are terminals. A cluster with only one terminal has no counted trifurcation. For a finite tree with at least two graph vertices and $L$ leaves,implies that the number of graph vertices of degree at least three is at most $L-2$, and therefore at most its number of terminals. Distinct exterior components can be assigned distinct graph vertices of ${\partial }^{+}W$. Terminals from different clusters are disjoint as well. Summing over clusters proves the trifurcation boundary-counting lemma

Take $W=[-n,n{]}^d\cap {\mathbb{Z}}^d$. The expected left side is $q|W|$, of order $q{n}^d$, while the right side is $O_d(n^{d-1})$. Division by $|W|$ and passage to infinity force $q=0$, contradicting the positive probability obtained by finite modification of Bernoulli percolation. Thus infinitely many infinite percolation clusters are impossible. The surface-to-volume comparison is the geometric reason for uniqueness on the lattice; the same argument is tied to amenability on more general graphs.

At $p=0$ there is no infinite percolation cluster, and at $p=1$ the whole lattice is the unique infinite percolation cluster. This completes all parameter cases. Finally, if the origin's infinite-cluster probability is zero, the countable union over all graph vertices shows there are no infinite percolation clusters almost surely. If it is positive, the translation-invariant existence event has positive probability and hence probability one. Together with uniqueness this distinguishes absence from the case of a single infinite percolation cluster, without making an unsupported assertion that every graph vertex belongs to it.